Problem 26 - College Physics

Question

A slit of width \(0.15 \mathrm{~mm}\) is illuminated with monochromatic light of wavelength \(632.8 \mathrm{nm}\). At what angle will the first minimum occur?

Step-by-Step Solution

Answer

The first minimum occurs at approximately 0.241 degrees.

Step 1: Understand the Question

We are given a slit of width 0.15 mm, and it is illuminated with monochromatic light having a wavelength of 632.8 nm. We are tasked with finding the angle at which the first diffraction minimum will occur.

Step 2: Identify Relevant Formula

The angle for the first minimum in a single-slit diffraction pattern can be found using the formula: \( a \sin\theta = m\lambda \), where \( a \) is the slit width, \( \theta \) is the diffraction angle, \( \lambda \) is the wavelength, and \( m \) is the order of the minimum (we use \( m=1 \) for the first minimum).

Step 3: Convert Units

First, convert the given slit width and wavelength to compatible units. The slit width \( a = 0.15 \, \text{mm} = 0.15 \times 10^{-3} \, \text{m} \) and wavelength \( \lambda = 632.8 \, \text{nm} = 632.8 \times 10^{-9} \, \text{m} \).

Step 4: Apply the Formula

Substitute the values into the formula: \( 0.15 \times 10^{-3} \sin \theta = 1 \times 632.8 \times 10^{-9} \).

Step 5: Solve for Angle \( \theta \)

Rearrange the equation to solve for \( \sin \theta \):\[ \sin \theta = \frac{632.8 \times 10^{-9}}{0.15 \times 10^{-3}} = \frac{632.8 \times 10^{-6}}{0.15} \]Calculate the value: \( \sin \theta \approx 0.004219 \).

Step 6: Calculate \( \theta \)

Take the inverse sine of both sides to find \( \theta \):\[ \theta = \sin^{-1}(0.004219) \]Using a calculator, you find \( \theta \approx 0.241 \) degrees.

Key Concepts

Monochromatic Light

Monochromatic light refers to light that consists of a single wavelength. It means that the light waves are all of the same color or frequency. This simplicity makes it ideal for experiments in physics, such as single-slit diffraction studies. In these settings, monochromatic light sources can be lasers, which produce a consistent and stable wavelength. Utilizing monochromatic light helps to ensure that any observed phenomena are due to the experimental setup without interference from other wavelengths.
In the context of the exercise, we're dealing with light of wavelength 632.8 nm. Since one nanometer (nm) is equal to one billionth of a meter, this means our light is in the red part of the visible spectrum. This specific wavelength underscores the concept's importance in accurately predicting where patterns like minima and maxima in diffraction occur.

Diffraction Minimum

In optics, a diffraction minimum is a point where light intensity falls to zero due to destructive interference. This occurs when waves originating from different parts of the slit cancel each other out. In a single-slit diffraction experiment, minima are found at specific angles where this cancellation happens.
The condition for the first diffraction minimum can be described using the formula:

• \( a \sin \theta = m\lambda \)

where \( a \) is the slit width, \( \theta \) is the angle of the minimum, \( \lambda \) is the wavelength, and \( m \) is the order number of the minimum. For the first minimum, \( m = 1 \). This formula is essential because it provides a quantitative way to find these angles, which are critical in understanding and analyzing diffraction patterns.

Wavelength Conversion

Since physics equations often require consistent units, converting the measured values into compatible units is crucial. In the original problem, we have the slit width given as 0.15 mm and the wavelength as 632.8 nm. To ensure accurate calculations, both need to be converted to meters, the standard SI unit for length.
This conversion involves:

• Converting millimeters to meters: \(0.15 \, \text{mm} = 0.15 \times 10^{-3} \, \text{m}\).
• Converting nanometers to meters: \(632.8 \, \text{nm} = 632.8 \times 10^{-9} \, \text{m}\).

By maintaining consistent units, we can apply mathematical formulas without computational errors, ensuring the integrity of the solution.

Angle Calculation

Angle calculation in single-slit diffraction is about determining at which angle a specific phenomenon, such as a diffraction minimum, occurs. For the given problem, we use the equation:

• \( \sin \theta = \frac{m\lambda}{a} \)

With known slit width \( a \) and wavelength \( \lambda \), this equation helps calculate \( \theta \), the angle of the diffraction minimum. First, solve for \( \sin \theta \):\( \sin \theta = \frac{632.8 \times 10^{-9}}{0.15 \times 10^{-3}} = 0.004219 \).
Next, find \( \theta \) using the inverse sin function: \( \theta = \sin^{-1}(0.004219) \). After calculation, you'll find \( \theta \approx 0.241 \) degrees. This process determines the precise angle at which the first minimum occurs, illustrating the critical role of precise calculation in understanding diffraction patterns.